Ferrari-Cardano derivation of the quartic formula

Clearly a solution to Equation (1) solves the original, so we replace the original equation with Equation (1). Move q⁢y+r to the other side and complete the square on the left to get:

(y2+p)2=p⁢y2-q⁢y+(p2-r).

We now wish to add the quantity (y2+p+z)2-(y2+p)2 to both sides, for some unspecified value of z whose purpose will be made clear in what follows. Note that (y2+p+z)2-(y2+p)2 is a quadratic in y. Carrying out this addition, we get

Our goal will be achieved if we can find a value for z which makes this discriminant zero. But the above polynomial is a cubic polynomial in z, so its roots can be found using the cubic formula. Choosing then such a value for z, we may rewrite Equation (2) as

(y2+p+z)2=(s⁢y+t)2

for some (complicated!) values s and t, and then taking the square root of both sides and solving the resulting quadratic equation in y provides a root of Equation (1).